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NCERT Solutions · Class 12 Mathematics Probability

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Miscellaneous Exercise 1–13 (part 7 of 7)

  1. Exercise 1

    A and B are two events such that P(A)≠0\displaystyle \mathrm{P}(\mathrm{A}) \neq 0. Find P(B∣A)\displaystyle \mathrm{P}(\mathrm{B} \mid \mathrm{A}), if
    (i)
    A is a subset of B
    (ii)
    A∩B=ϕ\displaystyle \mathrm{A} \cap \mathrm{B}=\phi

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    NCERT’s answer
    (i)
    $\displaystyle 1$ (ii) $\displaystyle 0$
    By the definition of conditional probability, since \(\displaystyle \mathrm{P}(\mathrm{A})\neq 0\),
    \[\mathrm{P}(\mathrm{B}\mid \mathrm{A})=\frac{\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}. \]
    So everything turns on what \(\displaystyle \mathrm{A}\cap \mathrm{B}\) is in each case.
    (i)
    \(\displaystyle \mathrm{A}\) is a subset of \(\displaystyle \mathrm{B}\). Then every outcome favourable to \(\displaystyle \mathrm{A}\) is also favourable to \(\displaystyle \mathrm{B}\), so \(\displaystyle \mathrm{A}\cap \mathrm{B}=\mathrm{A}\) and therefore \(\displaystyle \mathrm{P}(\mathrm{A}\cap \mathrm{B})=\mathrm{P}(\mathrm{A})\). Hence
    \[\mathrm{P}(\mathrm{B}\mid \mathrm{A})=\frac{\mathrm{P}(\mathrm{A})}{\mathrm{P}(\mathrm{A})}=1. \]
    (This is the expected reading: once \(\displaystyle \mathrm{A}\) has occurred, \(\displaystyle \mathrm{B}\) is certain.)
    (ii)
    \(\displaystyle \mathrm{A}\cap \mathrm{B}=\phi\). The events are mutually exclusive, so \(\displaystyle \mathrm{P}(\mathrm{A}\cap \mathrm{B})=\mathrm{P}(\phi)=0\), and
    \[\mathrm{P}(\mathrm{B}\mid \mathrm{A})=\frac{0}{\mathrm{P}(\mathrm{A})}=0. \]
    Final answer: (i) \(\displaystyle \mathrm{P}(\mathrm{B}\mid \mathrm{A})=1\); (ii) \(\displaystyle \mathrm{P}(\mathrm{B}\mid \mathrm{A})=0\).

  2. Exercise 2

    A couple has two children,
    (i)
    Find the probability that both children are males, if it is known that at least one of the children is male.
    (ii)
    Find the probability that both children are females, if it is known that the elder child is a female.

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    NCERT’s answer
    (i)
    \(\displaystyle \frac{1}{3}\) (ii) \(\displaystyle \frac{1}{2}\)
    Write each outcome as an ordered pair (elder child, younger child), with \(\displaystyle b\) for a boy and \(\displaystyle g\) for a girl. The sample space is
    \[\mathrm{S}=\{(b,b),\ (b,g),\ (g,b),\ (g,g)\}, \]
    four equally likely outcomes, each of probability \(\displaystyle \tfrac14\). Note that the order matters here: \(\displaystyle (b,g)\) and \(\displaystyle (g,b)\) are different outcomes, and forgetting this is what makes part (i) come out wrong.
    (i)
    Let \(\displaystyle \mathrm{E}=\) "both children are males" \(\displaystyle =\{(b,b)\}\) and \(\displaystyle \mathrm{F}=\) "at least one child is male" \(\displaystyle =\{(b,b),(b,g),(g,b)\}\). Then \(\displaystyle \mathrm{E}\cap \mathrm{F}=\{(b,b)\}\), so
    \[\mathrm{P}(\mathrm{E}\mid \mathrm{F})=\frac{\mathrm{P}(\mathrm{E}\cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}=\frac{1/4}{3/4}=\frac13. \]
    (ii)
    Let \(\displaystyle \mathrm{E}=\) "both children are females" \(\displaystyle =\{(g,g)\}\) and \(\displaystyle \mathrm{F}=\) "the elder child is a female" \(\displaystyle =\{(g,b),(g,g)\}\). Then \(\displaystyle \mathrm{E}\cap \mathrm{F}=\{(g,g)\}\), so
    \[\mathrm{P}(\mathrm{E}\mid \mathrm{F})=\frac{1/4}{2/4}=\frac12. \]
    Final answer: (i) \(\displaystyle \dfrac13\); (ii) \(\displaystyle \dfrac12\).

  3. Exercise 3

    Suppose that 5\displaystyle 5% of men and 0.25\displaystyle 0.25% of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male? Assume that there are equal number of males and females.

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    NCERT’s answer
    \(\displaystyle \frac{20}{21}\)
    Let \(\displaystyle \mathrm{E}_1=\) "the person chosen is male", \(\displaystyle \mathrm{E}_2=\) "the person chosen is female", and \(\displaystyle \mathrm{A}=\) "the person has grey hair". Since males and females are equal in number, \[\mathrm{P}(\mathrm{E}_1)=\mathrm{P}(\mathrm{E}_2)=\frac12 . \] The given percentages are conditional probabilities: \[\mathrm{P}(\mathrm{A}\mid \mathrm{E}_1)=5\%=\frac{5}{100},\qquad \mathrm{P}(\mathrm{A}\mid \mathrm{E}_2)=0.25\%=\frac{0.25}{100}. \] We want \(\displaystyle \mathrm{P}(\mathrm{E}_1\mid \mathrm{A})\), so we use Bayes' theorem: \[\mathrm{P}(\mathrm{E}_1\mid \mathrm{A})=\frac{\mathrm{P}(\mathrm{E}_1)\,\mathrm{P}(\mathrm{A}\mid \mathrm{E}_1)}{\mathrm{P}(\mathrm{E}_1)\,\mathrm{P}(\mathrm{A}\mid \mathrm{E}_1)+\mathrm{P}(\mathrm{E}_2)\,\mathrm{P}(\mathrm{A}\mid \mathrm{E}_2)}. \] Substituting (keep \(\displaystyle 0.25\%=0.0025\), not \(\displaystyle 0.25\) — that is the usual slip): \[\mathrm{P}(\mathrm{E}_1\mid \mathrm{A})=\frac{\frac12\times 0.05}{\frac12\times 0.05+\frac12\times 0.0025}=\frac{0.05}{0.0525}=\frac{500}{525}=\frac{20}{21}. \]Final answer: the probability that the grey-haired person is male is \(\displaystyle \dfrac{20}{21}\).

  4. Exercise 4

    Suppose that 90\displaystyle 90% of people are right-handed. What is the probability that at most 6\displaystyle 6 of a random sample of 10\displaystyle 10 people are right-handed?

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    NCERT’s answer
    \(\displaystyle 1-\sum_{r=7}^{10}{ }^{10} \mathrm{C}_{r}(0.9)^{r}(0.1)^{10-r}\)
    Each person independently is right-handed with probability \(\displaystyle p=0.9\), so if \(\displaystyle \mathrm{X}\) is the number of right-handed people among \(\displaystyle n=10\), then \(\displaystyle \mathrm{X}\) is a binomial variate with \[\mathrm{P}(\mathrm{X}=r)={}^{10}\mathrm{C}_r\,(0.9)^r\,(0.1)^{10-r},\qquad r=0,1,\dots,10. \] "At most $\displaystyle 6$" means \(\displaystyle \mathrm{X}\le 6\). Summing seven terms directly is heavy, so use the complement \(\displaystyle \mathrm{P}(\mathrm{X}\le 6)=1-\mathrm{P}(\mathrm{X}\ge 7)\): \[\mathrm{P}(\mathrm{X}\le 6)=1-\sum_{r=7}^{10}{}^{10}\mathrm{C}_r\left(\frac{9}{10}\right)^{\!r}\left(\frac{1}{10}\right)^{\!10-r}=1-\frac{1}{10^{10}}\sum_{r=7}^{10}{}^{10}\mathrm{C}_r\,9^{r}. \] Evaluating the four terms: \[{}^{10}\mathrm{C}_7\,9^{7}=120\times 4782969=573956280, \] \[{}^{10}\mathrm{C}_8\,9^{8}=45\times 43046721=1937102445, \] \[{}^{10}\mathrm{C}_9\,9^{9}=10\times 387420489=3874204890, \] \[{}^{10}\mathrm{C}_{10}\,9^{10}=3486784401. \] Their sum is \(\displaystyle 9872048016\), so \[\mathrm{P}(\mathrm{X}\le 6)=1-\frac{9872048016}{10^{10}}=\frac{127951984}{10^{10}}\approx 0.0128. \]Final answer: \(\displaystyle \displaystyle \mathrm{P}(\mathrm{X}\le 6)=\sum_{r=0}^{6}{}^{10}\mathrm{C}_r(0.9)^r(0.1)^{10-r}=1-\frac{9872048016}{10^{10}}\approx 0.0128\).

  5. Exercise 5

    If a leap year is selected at random, what is the chance that it will contain 53\displaystyle 53 tuesdays?

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    NCERT’s answer
    \(\displaystyle \frac{2}{7}\)
    A leap year has \(\displaystyle 366\) days. Dividing by \(\displaystyle 7\), \[366=52\times 7+2, \] so a leap year is \(\displaystyle 52\) complete weeks plus \(\displaystyle 2\) extra consecutive days. The \(\displaystyle 52\) full weeks already supply \(\displaystyle 52\) Tuesdays, so the year has \(\displaystyle 53\) Tuesdays exactly when one of the two extra days is a Tuesday.The two extra days are consecutive, so the sample space of possibilities is \[\mathrm{S}=\{(\text{Sun,Mon}),(\text{Mon,Tue}),(\text{Tue,Wed}),(\text{Wed,Thu}),(\text{Thu,Fri}),(\text{Fri,Sat}),(\text{Sat,Sun})\}, \] which has \(\displaystyle 7\) equally likely outcomes (not \(\displaystyle 49\) — the second day is determined by the first).The favourable outcomes, those containing Tuesday, are \(\displaystyle (\text{Mon,Tue})\) and \(\displaystyle (\text{Tue,Wed})\): \(\displaystyle 2\) of them. Hence \[\mathrm{P}(53\ \text{Tuesdays})=\frac{2}{7}. \]Final answer: \(\displaystyle \dfrac{2}{7}\).

  6. Exercise 6

    Suppose we have four boxes A,B,C and D containing coloured marbles as given below:
    BoxMarble colour
    RedWhiteBlack
    A1\displaystyle 16\displaystyle 63\displaystyle 3
    B6\displaystyle 62\displaystyle 22\displaystyle 2
    C8\displaystyle 81\displaystyle 11\displaystyle 1
    D0\displaystyle 06\displaystyle 64\displaystyle 4
    One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A?, box B?, box C?

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    NCERT’s answer
    \(\displaystyle \frac{1}{15}, \frac{2}{5}, \frac{8}{15}\)
    Let \(\displaystyle \mathrm{E}_1,\mathrm{E}_2,\mathrm{E}_3,\mathrm{E}_4\) be the events that box A, B, C, D respectively is chosen, and let \(\displaystyle \mathrm{R}\) be the event that the marble drawn is red. The box is chosen at random from four boxes, so \[\mathrm{P}(\mathrm{E}_1)=\mathrm{P}(\mathrm{E}_2)=\mathrm{P}(\mathrm{E}_3)=\mathrm{P}(\mathrm{E}_4)=\frac14 . \] Each box holds \(\displaystyle 10\) marbles \(\displaystyle (1+6+3=6+2+2=8+1+1=0+6+4=10)\), so \[\mathrm{P}(\mathrm{R}\mid \mathrm{E}_1)=\frac{1}{10},\quad \mathrm{P}(\mathrm{R}\mid \mathrm{E}_2)=\frac{6}{10},\quad \mathrm{P}(\mathrm{R}\mid \mathrm{E}_3)=\frac{8}{10},\quad \mathrm{P}(\mathrm{R}\mid \mathrm{E}_4)=\frac{0}{10}=0. \] By the theorem of total probability, \[\mathrm{P}(\mathrm{R})=\frac14\left(\frac{1}{10}+\frac{6}{10}+\frac{8}{10}+0\right)=\frac14\cdot\frac{15}{10}=\frac{3}{8}. \] Now apply Bayes' theorem, \(\displaystyle \mathrm{P}(\mathrm{E}_i\mid \mathrm{R})=\dfrac{\mathrm{P}(\mathrm{E}_i)\mathrm{P}(\mathrm{R}\mid \mathrm{E}_i)}{\mathrm{P}(\mathrm{R})}\): \[\mathrm{P}(\mathrm{E}_1\mid \mathrm{R})=\frac{\frac14\cdot\dfrac{1}{10}}{\frac38}=\frac{1/40}{3/8}=\frac{1}{15}, \] \[\mathrm{P}(\mathrm{E}_2\mid \mathrm{R})=\frac{\frac14\cdot\dfrac{6}{10}}{\frac38}=\frac{6/40}{3/8}=\frac{6}{15}=\frac{2}{5}, \] \[\mathrm{P}(\mathrm{E}_3\mid \mathrm{R})=\frac{\frac14\cdot\dfrac{8}{10}}{\frac38}=\frac{8/40}{3/8}=\frac{8}{15}. \] Check: \(\displaystyle \frac{1}{15}+\frac{6}{15}+\frac{8}{15}=1\), as it must be, since box D contains no red marble and so contributes probability \(\displaystyle 0\).Final answer: \(\displaystyle \mathrm{P}(\text{box A}\mid \text{red})=\dfrac{1}{15}\), \(\displaystyle \mathrm{P}(\text{box B}\mid \text{red})=\dfrac{2}{5}\), \(\displaystyle \mathrm{P}(\text{box C}\mid \text{red})=\dfrac{8}{15}\).

  7. Exercise 7

    Assume that the chances of a patient having a heart attack is 40\displaystyle 40%. It is also assumed that a meditation and yoga course reduce the risk of heart attack by 30\displaystyle 30% and prescription of certain drug reduces its chances by 25\displaystyle 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?

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    NCERT’s answer
    \(\displaystyle \frac{14}{29}\)
    Let \(\displaystyle \mathrm{E}_1=\) "the patient follows the meditation and yoga course", \(\displaystyle \mathrm{E}_2=\) "the patient takes the drug", and \(\displaystyle \mathrm{A}=\) "the patient suffers a heart attack". The patient chooses one option of the two with equal probability, so \[\mathrm{P}(\mathrm{E}_1)=\mathrm{P}(\mathrm{E}_2)=\frac12 . \] The base chance of an attack is \(\displaystyle 0.40\). The percentages given are reductions of that risk, so they must be applied multiplicatively to \(\displaystyle 0.40\) (not subtracted as percentage points): \[\mathrm{P}(\mathrm{A}\mid \mathrm{E}_1)=0.40\times(1-0.30)=0.40\times 0.70=0.28, \] \[\mathrm{P}(\mathrm{A}\mid \mathrm{E}_2)=0.40\times(1-0.25)=0.40\times 0.75=0.30. \] By Bayes' theorem, \[\mathrm{P}(\mathrm{E}_1\mid \mathrm{A})=\frac{\mathrm{P}(\mathrm{E}_1)\mathrm{P}(\mathrm{A}\mid \mathrm{E}_1)}{\mathrm{P}(\mathrm{E}_1)\mathrm{P}(\mathrm{A}\mid \mathrm{E}_1)+\mathrm{P}(\mathrm{E}_2)\mathrm{P}(\mathrm{A}\mid \mathrm{E}_2)} =\frac{\frac12\times 0.28}{\frac12\times 0.28+\frac12\times 0.30} =\frac{0.28}{0.58}=\frac{28}{58}=\frac{14}{29}. \]Final answer: the probability that the patient followed the meditation-and-yoga course is \(\displaystyle \dfrac{14}{29}\).

  8. Exercise 8

    If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability 12\displaystyle \frac{1}{2} ).

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    NCERT’s answer
    \(\displaystyle \frac{3}{16}\)
    Write the determinant as \[\Delta=\left|\begin{array}{rr} a & b \\ c & d \end{array}\right|=ad-bc, \] where each of \(\displaystyle a,b,c,d\) is \(\displaystyle 0\) or \(\displaystyle 1\), chosen independently with probability \(\displaystyle \tfrac12\) each. The four entries therefore give \[n(\mathrm{S})=2\times 2\times 2\times 2=16 \] equally likely determinants, each of probability \(\displaystyle \tfrac{1}{16}\).Since \(\displaystyle a,b,c,d\in\{0,1\}\), both products \(\displaystyle ad\) and \(\displaystyle bc\) are \(\displaystyle 0\) or \(\displaystyle 1\). So \(\displaystyle ad-bc>0\) is possible only in the single case \[ad=1\quad\text{and}\quad bc=0, \] which then gives \(\displaystyle \Delta=1\).\(\displaystyle ad=1\) forces \(\displaystyle a=1\) and \(\displaystyle d=1\): exactly \(\displaystyle 1\) choice of the pair \(\displaystyle (a,d)\). \(\displaystyle bc=0\) means at least one of \(\displaystyle b,c\) is \(\displaystyle 0\): the pairs \(\displaystyle (b,c)=(0,0),(0,1),(1,0)\), that is \(\displaystyle 3\) choices out of \(\displaystyle 4\).Hence the number of favourable determinants is \(\displaystyle 1\times 3=3\), namely \[\left|\begin{array}{rr}1&0\\0&1\end{array}\right|,\qquad \left|\begin{array}{rr}1&0\\1&1\end{array}\right|,\qquad \left|\begin{array}{rr}1&1\\0&1\end{array}\right|, \] and \[\mathrm{P}(\Delta>0)=\frac{3}{16}. \]Final answer: \(\displaystyle \dfrac{3}{16}\).

  9. Exercise 9

    An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known: P( A fails )=0.2P( B fails alone )=0.15P( A and B fail )=0.15\begin{aligned} \mathrm{P}(\mathrm{~A} \text { fails }) & =0.2 \\ \mathrm{P}(\mathrm{~B} \text { fails alone }) & =0.15 \\ \mathrm{P}(\mathrm{~A} \text { and } \mathrm{B} \text { fail }) & =0.15 \end{aligned} Evaluate the following probabilities
    (i)
    P (A failsIB has failed)
    (ii)
    P (A fails alone)

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    NCERT’s answer
    (i)
    0.$\displaystyle 5$ (ii) $\displaystyle 0.05$
    Let \(\displaystyle \mathrm{A}\) be the event "subsystem A fails" and \(\displaystyle \mathrm{B}\) the event "subsystem B fails". We are given
    \[\mathrm{P}(\mathrm{A})=0.2,\qquad \mathrm{P}(\mathrm{B}\ \text{fails alone})=0.15,\qquad \mathrm{P}(\mathrm{A}\cap \mathrm{B})=0.15. \]
    "B fails alone" means B fails and A does not, i.e. the event \(\displaystyle \mathrm{B}\cap \mathrm{A}'\). Since \(\displaystyle \mathrm{B}=(\mathrm{B}\cap \mathrm{A})\cup(\mathrm{B}\cap \mathrm{A}')\) and these two parts are disjoint,
    \[\mathrm{P}(\mathrm{B}\ \text{fails alone})=\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A}\cap \mathrm{B}). \]
    This is the step to get right — \(\displaystyle \mathrm{P}(\mathrm{B})\) is not given directly. From it,
    \[0.15=\mathrm{P}(\mathrm{B})-0.15\ \Longrightarrow\ \mathrm{P}(\mathrm{B})=0.30. \]
    (i)
    By the definition of conditional probability,
    \[\mathrm{P}(\mathrm{A}\ \text{fails}\mid \mathrm{B}\ \text{has failed})=\frac{\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{0.15}{0.30}=0.5. \]
    (ii)
    By the same disjoint splitting applied to \(\displaystyle \mathrm{A}\),
    \[\mathrm{P}(\mathrm{A}\ \text{fails alone})=\mathrm{P}(\mathrm{A}\cap \mathrm{B}')=\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{A}\cap \mathrm{B})=0.2-0.15=0.05. \]
    Final answer: (i) \(\displaystyle 0.5\); (ii) \(\displaystyle 0.05\).

  10. Exercise 10

    Bag I contains 3\displaystyle 3 red and 4\displaystyle 4 black balls and Bag II contains 4\displaystyle 4 red and 5\displaystyle 5 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.

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    NCERT’s answer
    \(\displaystyle \frac{16}{31}\)
    Let \(\displaystyle \mathrm{E}_1=\) "a red ball is transferred from Bag I", \(\displaystyle \mathrm{E}_2=\) "a black ball is transferred from Bag I", and \(\displaystyle \mathrm{R}=\) "the ball drawn from Bag II is red".Bag I has \(\displaystyle 3\) red and \(\displaystyle 4\) black balls, i.e. \(\displaystyle 7\) in all, so \[\mathrm{P}(\mathrm{E}_1)=\frac37,\qquad \mathrm{P}(\mathrm{E}_2)=\frac47 . \] Bag II starts with \(\displaystyle 4\) red and \(\displaystyle 5\) black balls; after the transfer it holds \(\displaystyle 10\) balls, and which colour was added changes its composition: \[\text{if red was transferred: } 5\ \text{red},\ 5\ \text{black}\ \Rightarrow\ \mathrm{P}(\mathrm{R}\mid \mathrm{E}_1)=\frac{5}{10}=\frac12, \] \[\text{if black was transferred: } 4\ \text{red},\ 6\ \text{black}\ \Rightarrow\ \mathrm{P}(\mathrm{R}\mid \mathrm{E}_2)=\frac{4}{10}=\frac25 . \] By the theorem of total probability, \[\mathrm{P}(\mathrm{R})=\frac37\cdot\frac12+\frac47\cdot\frac25=\frac{3}{14}+\frac{8}{35}=\frac{15}{70}+\frac{16}{70}=\frac{31}{70}. \] We need the probability that the transferred ball was black, given the drawn ball is red, so by Bayes' theorem \[\mathrm{P}(\mathrm{E}_2\mid \mathrm{R})=\frac{\mathrm{P}(\mathrm{E}_2)\mathrm{P}(\mathrm{R}\mid \mathrm{E}_2)}{\mathrm{P}(\mathrm{R})}=\frac{\frac47\cdot\frac25}{\dfrac{31}{70}}=\frac{8/35}{31/70}=\frac{8}{35}\times\frac{70}{31}=\frac{16}{31}. \]Final answer: \(\displaystyle \dfrac{16}{31}\).
  11. Choose the correct answer in each of the following:

    Exercise 11

    If A and B are two events such that P(A)≠0\displaystyle \mathrm{P}(\mathrm{A}) \neq 0 and P(B∣A)=1\displaystyle \mathrm{P}(\mathrm{B} \mid \mathrm{A})=1, then (A) A⊂B\displaystyle \mathrm{A} \subset \mathrm{B} (B) B⊂A\displaystyle \mathrm{B} \subset \mathrm{A} (C) B=ϕ\displaystyle \mathrm{B}=\phi (D) A=ϕ\displaystyle \mathrm{A}=\phi

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    NCERT’s answer
    A
    Given \(\displaystyle \mathrm{P}(\mathrm{A})\neq 0\) and \(\displaystyle \mathrm{P}(\mathrm{B}\mid \mathrm{A})=1\). By the definition of conditional probability, \[\mathrm{P}(\mathrm{B}\mid \mathrm{A})=\frac{\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}=1\ \Longrightarrow\ \mathrm{P}(\mathrm{A}\cap \mathrm{B})=\mathrm{P}(\mathrm{A}). \] Since \(\displaystyle \mathrm{A}\cap \mathrm{B}\subseteq \mathrm{A}\) always, equality of their probabilities says that no part of \(\displaystyle \mathrm{A}\) lies outside \(\displaystyle \mathrm{B}\); indeed \[\mathrm{P}(\mathrm{A}\cap \mathrm{B}')=\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{A}\cap \mathrm{B})=0, \] so \(\displaystyle \mathrm{A}\subset \mathrm{B}\).The other options fail: \(\displaystyle \mathrm{B}\subset \mathrm{A}\) does not follow, \(\displaystyle \mathrm{B}=\phi\) would give \(\displaystyle \mathrm{P}(\mathrm{B}\mid \mathrm{A})=0\), and \(\displaystyle \mathrm{A}=\phi\) is ruled out by \(\displaystyle \mathrm{P}(\mathrm{A})\neq 0\).Final answer: option (A), \(\displaystyle \mathrm{A}\subset \mathrm{B}\).

  12. Exercise 12

    If P(A∣B)>P(A)\displaystyle \mathrm{P}(\mathrm{A} \mid \mathrm{B})>\mathrm{P}(\mathrm{A}), then which of the following is correct : (A) P(B∣A)<P(B)\displaystyle \mathrm{P}(\mathrm{B} \mid \mathrm{A})<\mathrm{P}(\mathrm{B}) (B) P(A∩B)<P(A)⋅P(B)\displaystyle \mathrm{P}(\mathrm{A} \cap \mathrm{B})<\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B}) (C) P(B∣A)>P(B)\displaystyle \mathrm{P}(\mathrm{B} \mid \mathrm{A})>\mathrm{P}(\mathrm{B}) (D) P(B∣A)=P(B)\displaystyle \mathrm{P}(\mathrm{B} \mid \mathrm{A})=\mathrm{P}(\mathrm{B})

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    NCERT’s answer
    C
    Given \(\displaystyle \mathrm{P}(\mathrm{A}\mid \mathrm{B})>\mathrm{P}(\mathrm{A})\), where implicitly \(\displaystyle \mathrm{P}(\mathrm{A})>0\) and \(\displaystyle \mathrm{P}(\mathrm{B})>0\). Using the definition of conditional probability and multiplying through by the positive number \(\displaystyle \mathrm{P}(\mathrm{B})\) (the inequality sign is preserved): \[\frac{\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}>\mathrm{P}(\mathrm{A})\ \Longrightarrow\ \mathrm{P}(\mathrm{A}\cap \mathrm{B})>\mathrm{P}(\mathrm{A})\,\mathrm{P}(\mathrm{B}). \] This already kills option (B). Now divide both sides by the positive number \(\displaystyle \mathrm{P}(\mathrm{A})\): \[\frac{\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}>\mathrm{P}(\mathrm{B})\ \Longrightarrow\ \mathrm{P}(\mathrm{B}\mid \mathrm{A})>\mathrm{P}(\mathrm{B}), \] which contradicts (A) and (D) and establishes (C).Final answer: option (C), \(\displaystyle \mathrm{P}(\mathrm{B}\mid \mathrm{A})>\mathrm{P}(\mathrm{B})\).

  13. Exercise 13

    If A and B are any two events such that P(A) + P(B) - P(A and B) = P(A), then (A) P(B∣A)=1\displaystyle \mathrm{P}(\mathrm{B} \mid \mathrm{A})=1 (B) P(A∣B)=1\displaystyle \mathrm{P}(\mathrm{A} \mid \mathrm{B})=1 (C) P(B∣A)=0\displaystyle \mathrm{P}(\mathrm{B} \mid \mathrm{A})=0 (D) P(A∣B)=0\displaystyle \mathrm{P}(\mathrm{A} \mid \mathrm{B})=0

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    NCERT’s answer
    B \section*{Notes}
    The given relation is \[\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A}\cap \mathrm{B})=\mathrm{P}(\mathrm{A}). \] Cancelling \(\displaystyle \mathrm{P}(\mathrm{A})\) from both sides, \[\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A}\cap \mathrm{B})=0\ \Longrightarrow\ \mathrm{P}(\mathrm{A}\cap \mathrm{B})=\mathrm{P}(\mathrm{B}). \] (Equivalently, the left side is \(\displaystyle \mathrm{P}(\mathrm{A}\cup \mathrm{B})\) by the addition theorem, so \(\displaystyle \mathrm{P}(\mathrm{A}\cup \mathrm{B})=\mathrm{P}(\mathrm{A})\), meaning \(\displaystyle \mathrm{B}\) adds nothing outside \(\displaystyle \mathrm{A}\).)Hence, with \(\displaystyle \mathrm{P}(\mathrm{B})\neq 0\) so that the conditional probability is defined, \[\mathrm{P}(\mathrm{A}\mid \mathrm{B})=\frac{\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{\mathrm{P}(\mathrm{B})}{\mathrm{P}(\mathrm{B})}=1. \] Note it is \(\displaystyle \mathrm{P}(\mathrm{A}\mid \mathrm{B})\), not \(\displaystyle \mathrm{P}(\mathrm{B}\mid \mathrm{A})\), that must equal \(\displaystyle 1\) here: the condition says \(\displaystyle \mathrm{B}\subset \mathrm{A}\), so \(\displaystyle \mathrm{B}\) occurring guarantees \(\displaystyle \mathrm{A}\), but not the other way round.Final answer: option (B), \(\displaystyle \mathrm{P}(\mathrm{A}\mid \mathrm{B})=1\).